y = 2x^3 - 8x^2 + 10x - 4 is a function defin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the function $y = 2x^3 - 8x^2 + 10x - 4$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 6x^2 -

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(D) Given the function $y = 2x^3 - 8x^2 + 10x - 4$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 6x^2 - 16x + 10$. Since the tangent is parallel to the $X$-axis,its slope must be zero: $\frac{dy}{dx} = 0 \implies 6x^2 - 16x + 10 = 0$. Dividing by $2$,we get $3x^2 - 8x + 5 = 0$. Factoring the quadratic equation: $3x^2 - 3x - 5x + 5 = 0 \implies 3x(x - 1) - 5(x - 1) = 0$. $(3x - 5)(x - 1) = 0$. This gives $x = 1$ or $x = \frac{5}{3}$. Since the point $(a, b)$ satisfies $a \in (1, 2)$,we discard $x = 1$ and accept $a = \frac{5}{3}$.

$y = 2x^3 - 8x^2 + 10x - 4$ is a function defined on $[1, 2]$. If the tangent drawn at a point $(a, b)$ on the graph of this function is parallel to the $X$-axis and $a \in (1, 2)$,then $a =$

Similar Questions

If $x + y = 16$ and $x^2 + y^2$ is minimum,then the values of $x$ and $y$ are

The condition that $f(x) = ax^3 + bx^2 + cx + d$ has no extreme value is

Let the radius and height of a right circular cylinder be related as $r^2 + h = 6$. If the volume of the cylinder is maximum,then the value of $\frac{r}{h}$ is:

The abscissae of the points of the curve $y = x(x - 2)(x - 4)$ where the tangents are parallel to the $x$-axis are obtained as:

The perimeter of a triangle is $10 \text{ cm}$. If one of its sides is $4 \text{ cm}$,then the remaining sides of the triangle,when the area of the triangle is maximum,are

Vedclass Test Series

Exam Paper Generator

Online Exam Module